Given $a_{n+1}=a_{n}+\frac{1}{a_{n}}$, and $a_0=5,$ how can I find $\lfloor a_{1000} \rfloor$? I've tried coming up with reasonable bounds, but to no avail (ex. ones like $\sqrt{x^2+1} < x + \frac{1}{x}$ are terrible).
I have looked at the only similar post I can find on this site (Solving $ a(n+1) = a(n) + \frac{1}{a(n)}$ with $a(1) = 1 $), but it doesn't seem to help.
Thanks in advance!